Determine the Equation Of Sinusodial Function

**mmhaaz** · October 19th 2009, 03:39 PM

Could someone please explain how I would find the ht of sine and cosine on this graph? I don't understand what the difference is. Also how would I find the period? Is it the same for cosine and sine? Thank-you to anyone that could help!! Not in radian form please...

**Grandad** · October 20th 2009, 06:15 AM

Hello mmhaaz

Originally Posted by mmhaaz

Could someone please explain how I would find the ht of sine and cosine on this graph? I don't understand what the difference is. Also how would I find the period? Is it the same for cosine and sine? Thank-you to anyone that could help!! Not in radian form please...

First: the difference between the graphs of sine and cosine: there is none, except a 'phase shift' of $90^o$ . In other words, if you shift the cosine graph $90^o$ to the right, you get the sine graph. See the attached diagram.

Note in particular the 'starting' values; i.e. the values when $x = 0$ :

$\cos 0^o=1$
$\sin0^o=0$

These will help you remember which graph is which.

Then, the particular questions relating to the graph you've attached.

The scale on the vertical axis isn't very clear, but it looks as if one square = one unit.

So, on that assumption, the graph has a maximum value of $1$ and a minimum value of $-5$ . The sinusoidal axis is the horizontal line that divides the graph exactly in half. So this is the line $y = -2$ (which I think is what you got).

Then, the amplitude is the vertical distance from the sinusoidal axis to one of the peaks or troughs. So that's $3$ (again, I think you got that).

From these two facts, we can deduce that the equation of the graph will be of the form:

$y=-2+3\sin$ (something) or $y = -2+3\cos$ (something)

The question then refers to the 'starting point', based on a sine and then a cosine graph. I assume that means the point which corresponds to the point where $x = 0$ on the graphs of the basic graphs (the ones in my diagram).

So, for sine, this is the point that corresponds to $(0,0)$ on the basic graph - in other words one of the points where the graph crosses the sinusoidal axis and is moving upwards in a direction from left to right. There's more than one such point, but the obvious one is $(30, -2)$ . (Can you see that another one will be $(210,-2)$ ?)

For the cosine function, the starting point will be the point corresponding to $(0,1)$ on the basic graph. So we're looking for a peak (maximum) value: the obvious one here is $(75,1)$ .

The period is the horizontal distance between two consecutive corresponding points on the graph. We said above that two possible starting points for the sine graph are $(30,-2)$ and $(210,-2)$ . So the period is the distance between these two points. That's $180^o$ .

Divide by $360^o$ for the next answer: $\frac{180}{360}=\frac12$ .

If we divide $x$ by this fraction ( $\tfrac12$ ) we get $2x$ , which gives us the coefficient of $x$ in the functions we're looking for. (A complete period for $y = \sin x$ is when $x$ goes from $0^o$ to $360^o$ ; so if we want a period of $180^o$ , we shall need values of $2x$ , because they will go from $0^o$ to $360^o$ when $x$ goes from $0^o$ to $180^o$ .)

The 'starting points' that we've worked out above tell us the angle that has to be added or subtracted from $x$ to give the correct starting value - the phase-shift.

So, for the sine function we want to start at $x = 30$ , so the function is